\newproblem{lay:4_5_20}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.5.20}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Jan. 20th, 2014} \\}{}

  % Problem statement
	Mark each statement as true or false. Justify each answer.
	\begin{enumerate}
		\item $\mathbb{R}^2$ is a two-dimensional subspace of $\mathbb{R}^3$
		\item The number of variables in the equation $A\mathbf{x}=\mathbf{0}$ equals the dimension of $\mathrm{Nul}\{A\}$.
		\item A vector space is infinite-dimensional if it is spanned by an infinite set.
		\item If $\mathrm{dim}\{V\}=n$ and if $S$ spans $V$, then $S$ is a basis of $V$.
		\item The only three-dimensional subspace of $\mathbb{R}^3$ is $\mathbb{R}^3$ itself.
	\end{enumerate}
}{
  % Solution
	\begin{enumerate}
		\item False, $\mathbb{R}^2$ is not a subpace of $\mathbb{R}^3$. Although it is true that $\mathbb{R}^2$ is isomorph to a two-dimensional
		      subspace of $\mathbb{R}^3$.
		\item False, the number of variables in the equation $A\mathbf{x}=\mathbf{0}$ equals the number of columns of $A$. The dimension of $\mathrm{Nul}\{A\}$ is
		      equal to the number of free variables in the equation $A\mathbf{x}=\mathbf{0}$.
		\item False, the infinite set must be a basis of $A$, that is, there cannot be a proper subset of $S$ that also spans the same vector space.
		\item False, for this to be true we also need that $S$ has $n$ vectors.
		\item True, $\mathbb{R}^3$ is said to be an improper subspace of itself.
	\end{enumerate}
}
\useproblem{lay:4_5_20}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
